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Why does flow separation cause a loss of lift, considering the low pressure is the cause of lift above the wing?
When the flow separates, the main stream of air is no longer passing by the upper surface of the wing. Usually this means that the air pressure will decrease above the wing, because the freestream flow can't 'get' to the area behind the wing (AKA the stalled area).

Something else necessary to this question :
When the flow turns downwards over the upper surface of a wing like you can see in the picture below, that doesn't necessarily make lift¹.

The low pressure created from this turning on top of the wing is what makes the lift¹. According to the last comment of this answer on Physics.SE.

The pressure difference creates the force (not the bending). The reasoning of why the pressure difference is created is more complex.

Back to the main question, now knowing these 2 things:

The turning of this flow doesn't create the lift, only the low pressure (which technically means that the turning does create the lift, but it does so by creating the low pressure)

There is low pressure created in the stalled area of wing

So if there is low pressure created in the stalled area of a wing anyways, and all that the flow turning downwards over the wing does is create low pressure, why does lift decrease when flow separates?
The way that I see it is that if the stalled area is low pressure, and all the flow turning downwards over the wing does is make low pressure, then why does the lift change if there is still low pressure above the wing in both cases?
(The actual magnitude of the low pressure might be different in a stall compared to the low pressure created from the flow turning downwards over the wing, which isn't something I considered)

So in case the question wasn't exactly obvious, why does the flow separating suddenly decrease the lift, given the information above?
¹: Even though I linked a source I am not 100% sure, although I see no other way for the flow turning to make lift, so I might be mistaken.

Does this answer your question?  https://aviation.stackexchange.com/questions/55163/why-does-a-stall-decrease-lift-rather-than-increasing-it?rq=1

@MichaelHall That top answer is actually what sparked this question. The answer states that the air that flows over the wing needs to get it's momentum vector pointed down slightly, but as stated in my question, the 'bending' of air doesn't create the lift (I don't think) so that doesn't really help.

@MichaelHall Also, do you know why this is getting downvoted? I provided sources, and some of the statements I knew I could be mistaken about I made a disclaimer at the bottom for. If I know what people think is wrong with this question I would be more than happy to edit it.

There are many questions here about lift and stalls.  Look up "downwash" and google Newton's 3rd law of motion.  The downvotes are likely for lack of any real effort.

@MichaelHall It might not reflect in my question, but I usually do quite a bit of research before asking, and that includes browsing this site. All of the other sources I found assumed the downward movement of air over the wing was  a cause of lift, however as stated with a source, that isn’t 100% accurate. Also, if you look at the help center, you’re supposed to downvote when there is a lack of research effort or the question doesn’t provide enough information to be answerable. In my question, I provided a source to back up my claim, and I believe that there was plenty of information.

Ok, just my opinion, but it feels like you are beating a dead horse...

@MichaelHall I can see where you’re coming from, but maybe when I figure out this question it will be obvious why it’s different.

Look at the first smoke trail over the airfoil, see how uniform, even and close it is to the surface? That is "laminar" flow and it places fast flowing air as close to the upper surface of the airfoil as you can get developing the greatest lift. Now consider that flow separated and turbulent mixing fast and slow air near the airfoil surface. the average velocity across the upper surface will be greatly reduced, reducing lift. Understand separation doesn't happen all-at-once. It will begin near the trailing edge and move forward, and from the wing-root out. (generally)

This is an interesting question. Perhaps a good way to break it down:

The local shape of the airfoil forces turning of the air packets
Due to the turning of the air packets, the local pressure decreases
The local pressure acts on the surface of the wing (and this is how the force is transferred from the air to the wing)
The sum of the forces on the surface combines to lift & drag.

If we look at the example of the cylinder, here for different Reynolds numbers.

Source
We then take a closer look at separated vs not-separted.
We therefore zoom in on the first two images. Here the first image has no separation, whereas the second has flow separation. Have a closer look at the blue and red streamlines, and then specifically in the area I marked with a circle.
You can see that the red line (the one that is separated) has much less curvature in that area than the blue one. As a consequence, the pressure in that area will not be as low as it is in the blue case.

Source: Edit of the above image
Flow separation on the wing has the same effect (see the images below), it reduces the curving of the air packets, thereby reducing the negative pressure, and reducing the amount of lift generated.

Source

I see. So the pressure of separated flow is high than that of attached flow, like you said with your example. If you use that logic, how does flow separation create more drag? If the pressure is higher with separated flow behind an object, why would it create more drag than attached flow?

(If that doesn’t make sense I can try to explain it better)

I think that's another question to ask. The direct relationship between curvature and pressure only holds for inviscid flows (but can be useful in viscous flows). Drag in separated flow is dominated by viscous effects (to understand why drag is larger, look at the size of the 'dead water' region in Peter's answer).

I see. So by viscous effects do you mean the turbulence ‘mixing’ the flow above the wing to where the wing has to transfer more of its energy to the surrounding air?

@Wyatt I would recommend you look at Fundamentals of Aerodynamics, by John. D. Anderson, specifically p.390-393. It talks about exactly this.
This book is a great reference to learn about the principles of aerodynamics, and it would be a great starting point for your learning journey. It is a great resource, with the information structured coherently, which I think will benefit you greatly.

Oh interesting, I actually have that book. I will take a look when I can.

To illustrate the effect of flow separation on lift, consider what the flow outside of the boundary layer "sees" when flowing around the wing. At low angle of attack flow is attached, the boundary layer stays thin and the flowlines follow the contour of the wing, give and take the few millimeters of boundary layer.
At high angle of attack the flow separates on the upper rear wing and the boundary layer grows massively, filled up by a turbulent deadwater which presents a quite different airfoil to the outside flow. Since the thickening only happens on the upper side, the effective angle of attack of this wing-plus-deadwater airfoil is quite a bit lower than the geometric angle of attack of the original airfoil
In the sketch below you will see the airfoil at two different angles of attack (grey), the flowlines around both (blue) and the boundary layer air (red). The difference should be easy to spot.

The wing with separated flow still produces considerable lift, but much less than what the pure wing would be capable of if the flow would be inviscid (so all the red stuff would be missing in those sketches).
BTW, that dead water will be "eaten up" by the outside flow by the same mechanism which created the boundary layer: Shear will eventually equalise the speed of all the air behind the wing.

Most of the lift force is from the large package of air that was induced to move down by the wing's passage, not so much from the "suction" of the LP area pulling the wing up. The low pressure area and its distribution profile, moving in a smooth non chaotic manner, is what induces this movement, extending several chords above the wing (it's why you put spoilers on top to kill lift).
When a wing stalls, the low pressure region is still there, but because it's turbulent, its ability to induce a large package of air above it to move down is destroyed, and the total mass of air being moved down is limited to air that is directly deflected by the wing itself (plus what pressure differential there is).

If you were to look at it on a molecular level where the LP area above the wing turns the flow: If the flow were not to turn, then there would be a complete vacuum. The flow does turn because the LP forces it to turn. On a molecular level, the area of more air molecules would travel into the LP area above the wing where there is less air molecules. It’s like the atmosphere pressure is bending the air over the wing, and not the low pressure. You could also look at it as low pressure never pulls anything. When I have more time, I will be explaining this better, in case it doesn’t make sense.

Not only the turbulence limits the contribution of the area to the downwash, but also the adverse velocity gradient close to the airfoil in the separation bubble actually opposes to the downwash.

@mins I don’t think you were intending to answer my whole question with that comment, but you just might’ve! So if I’m understanding you correctly, (or at least part of what you said) the turbulent air mixes the flow in the stalled area, so the pressure isn’t able to be as low as it would without the turbulence? Essentially, the low pressure air in the stalled area is forced out by the turbulence mixing the air.

@mins With a few of the other answers put together, I think I understand that part now. However, using the same logic that you just explained, why would separation cause a drag increase? I could definitely see the L/D ratio being changed, but why would the separation itself change the drag?

@mins “In the turbulent boundary layer the pressure gradient close to the airfoil is larger” - If you don’t mind answering, why is that the case? Why would turbulence change any pressure gradients?

@Wyatt: Turbulence changes velocities and introduces negative velocities (adverse gradient). To simplify, say pressure is linked to velocity (by Bernoulli's principle) also in the turbulent BL. Velocities in the two BL are shown here. Y is the height in the BL (0=wall, 1=outer limit of BL) and X is the fraction of free stream velocity U0 at this height. You see the steeper curve for the TBL (source). I suggest we continue in the chat for more clarifications.

@mins That actually does make sense, thanks. Here is the chat room that I just created.

So if there is low pressure created in the stalled area of a wing anyways, and all that the flow turning downwards over the wing does is create low pressure, why does lift decrease when flow separates?

The premise of your question is not correct.
The following picture (source) shows how on an airfoil the pressure distribution changes with the AoA:

According to the source, this particular airfoil stalls at 11° so that the plots in the middle and on the right are for stalled conditions.
As you can see, when the airfoil stalls the pressure on the upper surface (lower part of the plots) becomes flat and the peak of suction around its leading edge (seen on the left plot) just disappears.

I see. The air accelerating over the upper surface is what causes the suction in your picture. When you increase AoA, won’t the upper surface also become lower pressure because of the higher AoA regardless of the acceleration of air? See here to see what I am talking about.

Basically, the suction peak from the accelerating air can’t be greater than the high pressure that did the accelerating of that air. Then also, as the link says, the high pressure on the front of an object (the lower surface of an airfoil at high AoA) will have an equal but opposite pressure on the back of the object (the upper surface of the airfoil at high AoA).

If that didn’t help, essentially if you have an airfoil at high AoA to where the lower surface is contacting free stream air, the high pressure that creates will be mirrored by an equal but opposite low pressure on the other side.

"the high pressure on the lower surface of an airfoil at high AoA will have an equal but opposite pressure on the upper surface of the airfoil" not really since the flow on the upper\opposite side is fully detached and doesn't follow the shape of the airfoil. The high pressure on the side facing the flow wouldn't be compensated for low pressure on the shadowed side. Intuitively: if that were true then drag would be zero...

Consider the low pressure area in terms of the Bernoulli principle, thinking of the average speed of those molecules as they move from that point normal to the relative wind and the trailing edge. Averaged across the wing, turbulent air moves slower, with a corresponding higher pressure.
Assume smooth flow across the underside of the wing. A smooth flow across the upper surface would have a longer path, and motion parallel to the wing surface at any point. However, a turbulent flow contains some molecules moving away from the wing, and others angling toward the wing. It is all very messy or noisy: overall slower (with higher pressure) than when the airflow is smooth.

"the upper surface would have a longer path" length is totally unrelated to the problem, and this assertion is wrong for an airfoil without camber. Part of the equal transit time fallacy.

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center.


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How will a normal shock forming on a wing create drag?
When you're going close to supersonic speeds, the air over the wing is actually getting accelerated to supersonic. When the pressure recovery area comes, this accelerated air turns into a shock because it has to decelerate rapidly.
The thing is, all this normal shock is doing is decelerating the flow over the wing.
It would make sense if there was a normal shock on the front of an airfoil causing drag, because that would be pushing in the opposite direction of flight. However, the shock I'm talking about is on the rear section of the wing, and it travels towards the trailing edge as you speed up.
I don't really see how this normal shock could cause a drag increase knowing all it does it decelerate the flow, so how does this work?

(The only way I see this normal shock making drag is if it caused separation)

Your comment is largely correct -- the shock on the wing will almost always cause separation, which is a large source of drag.
This is the predominant cause of drag during the transonic drag rise.  It is often called 'wave drag', but that is erroneous.

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Garmin autopilot question
I recently took a check out in a Cessna 172 with a Garmin GFC 500. My instructor had said when the plane was turning on nav mode, to not select a different heading opposite of the turn and switch to heading mode. For example, the plane is making a right turn towards the east capturing a GPS track, is it a bad idea to select a north-northwest heading and switch from NAV mode to HDG? I don’t really see the problem with this.
Hopes this makes sense.

Why not ask the instructor what the reason is behind this guideline/procedure/technique/suggestion?

One reason that might be behind this is that it's maybe somewhat easier to inadvertently turn left when you intend to turn right or vice versa if the nav mode track and the new heading are close to 180º apart. But that's just a complete guess.

Well, it depends on what you consider to be a "problem".  The issue isn't unique to Garmin, most autopilots will immediately turn towards the heading or course of whatever source you have selected, so you just need to understand what will happen.
What will happen is this:   If the heading of the mode you switch to is less than 180° and opposite to the direction the airplane is already turning, it will reverse course in a turn to the new mode.  This reversal maybe a little bit rapid and startling if you aren't expecting it. Autopilots can be a bit jerky, and it could be disconcerting for passengers if your intent is to be smooth.
Even engaging heading from wings level can be jerky if the bug isn't at 12:00 on the HSI.
In the example you provided if your intent was to reverse your turn to a new heading to the north northwest, you'd probably want to disengage autopilot, level the wings, put the heading bug at the 12:00 position, and then engage heading mode. Then you could use the heading bug to smoothly turn the airplane to the new heading you want.
Alternately, if you wanted to leave autopilot engaged you could roll the heading bug 20 or 30° ahead of the direction of the turn, and then engage heading mode. The airplane will continue the right turn, then roll wings level when it gets to that heading.  Then you can gradually reverse the bug back the other direction to initiate a left turn to NNW.

Thank you for this information!!!! This does sound like it could be the answer. I’m meeting with the instructor tomorrow, and will definitely bring this up. I will update you on what he says about it.

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Why do planes use wire HF antennas instead of smaller ones used by airliners?
why do some planes have wire Hf antennas if smaller hub antennas are more aerodynamic? like on the Boeing 787? Why do some modern aircraft like the c130 still have them?

As a point of note, the versions of the C-130 which use wire antennas are far from modern! The J model which is the most modern (‘90s) doesn’t use a wire HF antenna.

The antenna length is purely a function of frequency.  In other words it is a function of the radio systems that were prevalent when the aircraft entered service, and not specifically the age of the aircraft and radios.
The wavelength of a radio signal at 4 MHz is about 80 meters or 250 feet long.  The wavelength of a radio signal in the typical VHF NAV/COM band is around 2 meters, or 6 feet.
There are certain lengths of antennas that are more efficient than others, and 1/4 wavelength and 5/8 wavelength antennae offer excellent tradeoffs between length and signal strength.  A typical NAV/COM "whip" on a light aircraft will often be about 18" long for exactly that reason.
More and more navigation is being done using much higher frequencies, especially as digital radio component costs have shrunk drastically.  Many of the services on your image are UHF or even above where the wavelengths involved can be 70cm, and GPS is around 20cm.  You're talking about quarter-wavelength antennae, or even phased arrays, that are a few inches across.
Today, ADF is the only long-wavelength radio service still sometimes used in instrument navigation, so you do sometimes see GA aircraft with the tail wire antennae for ADF sense.  But that is becoming increasingly rare as GPS becomes ubiquitous.
A notable place where you will see long wire antennae in active use?  Military, cargo, bush operations, etc.  And for two reasons:

Geographic:  There are still many places where you don't want to rely solely on GPS (more accurately GNSS) as your sole means of navigation, but there are no other sources of VHF or UHF navaids.  And so the backup to the GPS may very well be a HF beacon, an AM radio station, or even a regional shortwave broadcaster.

Communications:  There are other use cases that cannot be met by GPS.  In many military and remote area operations, that long wire is also a communication antenna on which HF comm signals are transmitted and received.  In military or remote operations, I have three choices:  Relatively short-range, line-of-sight VHF/UHF comms, satellite comms, and HF/LF comms capable of propagation over the horizon.  I'm not sure how common it is, but even recently the use of HF comms were common for oceanic airline flights,

I think the question is more interested in HF comm antennas -- some modern aircraft still have long-wire HF antennas, and others have the in-the-tail HF antennas.  Why are both still around, with one type used "here" but the other type used "there"?

That is the question. But I did like the answer given above.

@Boeing787 I may edit my answer to clarify then...

One thought I had was that you can’t hide an antenna in a metal fuselage but you could in a composite one.  Don’t know if that has any real bearing in it though

@Max so would you say that smaller HF antennas on airliners is not as powerful as wire antennas?

@Boeing787 It’s not quite that simple, as is so often the case.  Running 100 watts of transmit power into an antenna that has 3db of gain is the same effective radiated power as running 50 watts into an antenna with 6dB of gain.  The engineering objective is to provide reliable transmit and receive over the distance required, and as long as that objective is met, more is not necessarily any better.  And advances in digital signal processing have steadily shrunk the importance of antenna size over the last 30 or so years.

Replying to: I'm not sure how common it is, but even recently the use of HF comms were common for oceanic airline flights,
Yes it is common even mandatory (due to vast deserts) by some african countries aviation authorities to be at least one HF comm Tx/Rx fitted to the aircraft to be registered there.

@Max because I've seen pictures of Cessna citations and Pilatus pc12's with wire antennas.


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How does a thickening boundary layer create form drag?
Imagine a normal wing with flow over it, and to simplify things it's at 0 AoA. Drag is being made mainly because of form drag and skin friction drag. I'm not sure on this, but I don't think there would be any induced drag here.
To add to that, drag is any force that has a component facing opposite to the direction of flight.
In this answer, it says :

It is more correctly termed "profile drag" because the growth of the boundary layer virtually modifies the profile of the airfoil seen by the airflow.

This states that there is extra drag from the air in the growing boundary layer, and it is profile drag, or form drag.
The boundary layer air isn't connected to the wing, so why would it make (form) drag? Since the force of drag is from something 'pushing' backwards on a wing or body, why would faster moving flow (flow before it reaches airfoil) encountering slower moving flow (flow inside BL) make drag?
If the whole wing was increased to the size of the boundary layer it would make sense why form drag would increase, as there is more surface for the oncoming air to push against.
However I can't seem to see why air which isn't connected to the wing would create drag in the same way.

Profile drag is the term for skin friction and form drag on a 2D body.  We do not talk about profile drag on 3D bodies.
3D bodies still have skin friction and form drag.
In this picture, the thickness of the boundary layer is exaggerated.
Recall that the pressure drag around a 2D inviscid closed body will be zero.  This is because everything cancels out just right.
Next, we treat the flow outside the boundary layer as inviscid.  It produces a pressure distribution that pushes on the outside of the boundary layer.
Next, we assume that the pressure 'across' the boundary layer is constant (this goes with it being thin).  So, we apply the pressure distribution from the inviscid flow on the outside of the boundary layer to the airfoil shape.  We can integrate this pressure to get the lift.
If you really understand each of the above steps, you're ready to understand form drag.  If you aren't, then re-read these steps again and think about them.
What happens to the boundary layer at the end of the airfoil?
It keeps going.
The inviscid body (around which the pressures sum to zero drag) would need to continue on the outside of the boundary layer to infinity.
But we can't deal with a pressure distribution that goes to infinity.  So instead, we cut it off at the blue line I've added to this figure.
We don't apply the entire inviscid body's pressure distribution to the airfoil.  We just apply the part we chopped off and kept.
Which leaves us with the problem -- we needed a full closed body's worth of pressures to cancel out and get to zero.  What happens if we chop off a chunk and throw it away?  I.e. we aren't dealing with a closed body?
Another way to think about it is to integrate the pressure on the outside of the boundary layer (without transferring it through to the airfoil).  This integral would clearly need to extend to infinity to encompass the entire boundary layer.
Instead, when we cut it off at the blue line, and we do the integral around the BL near the airfoil -- but closed off by the blue line.  What pressure do we apply along the blue line?
Since pressure is constant across the boundary layer, it will be the pressure at the outside of the boundary layer at the trailing edge.
The blue line is nearly vertical -- which means that any force we integrate across it will be in the thrust-drag direction.
Also, we know the pressure on this face (the pressure at the TE of the boundary layer).  But we now see that the thicker the boundary layer, the longer the blue line.  Consequently, the thickness of the boundary layer at the trailing edge is directly related to the amount of form drag.
From all this, we see that form drag is a pressure drag -- but it is a pressure drag that only exists because of viscous effects (the presence of the boundary layer).

I tried finding out what a 2D inviscid closed body was (because I didn't really know), and after looking through google (reputable places) and Introduction to Flight, I couldn't find anything. I'm assuming that's essential to know to understand this answer?

Inviscid means without viscosity which is equivalent to ignore the boundary layer.

@AeroAndy Right, but I was more confused on what a 2d closed body is. It might be simpler than I'm thinking of. (An airfoil type shape)

yup, don't overthink it. just a two-dimensional shape with no openings. :)

Take a pencil and draw a loopy curve.  If you end the curve where you started it, it is closed (say a circle).  If you don't, it is open (like the letter C).  Two-dimensional just means something you can draw on a sheet of paper.

@RobMcDonald I see. So there are pressures acting on the blue line drawn. You might’ve already explained this, but where is that blue line in a real scenario? On a real wing or closed body, where is the force being applied; the same force as the blue line is experiencing?

That isn't always a simple answer.  In Engineering, we often draw what is called a control volume.  It is a dotted line that we draw around a system.  Sometimes we draw it tight up against the system.  Other times, we draw it far away from the system.  We then apply conservation laws (mass, momentum, energy) to the boundary of the control volume.  This helps us figure out what is going on with the control volume.  Because we know the conservation laws apply, we know that we can look a the pressure and momentum that goes through a control volume and we can solve the force that acts on it.

So, in addition to the pressure on the blue line, we have to keep track of the momentum flux across the blue line.  And of course, this depends on which definition of the boundary layer thickness you're using -- the 99% thickness, the momentum thickness, or the displacement thickness. These are details that are important in the formal derivations of these things -- we aren't talking about those details here.  One of the powerful things about using a control volume is that we can choose to draw the control volume in a way that makes the problem easier.  We don't have to figure out some details.


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Why do the shocks in supersonic jet exhaust end at the boundary?
As explained in this answer, the shocks in supersonic jet exhaust are created from the ambient air pressure compressing the exhaust stream. This eventually makes a shockwave, which reflects and then travels outwards to the boundary of the jet exhaust. My question is why doesn’t the shock just keep traveling outward instead stopping at the boundary to the atmospheric air as it does in the picture? (Look to the right of the red discs)

Notice where it says the nozzle is overexpanded -- Pe < Pa.  That means that at the Nozzle Exit, the pressure is less than ambient.
After the nozzle, there is no rigid wall against the flow, so the boundary between the jet exhaust (white) and the ambient (blue) is a slip line, which means pressure must be equal.
As the notes say, this forces the initial shock -- it must jump the pressure, but as a consequence, it also turns the flow.
There are a series of interactions and reflections that then happen.  It is often beneficial to think of the center line as a solid wall.  Symmetry can often be treated like a solid inviscid wall.
The white flow is supersonic -- usually substantially so.  The ambient flow is much slower.  Possibly supersonic, but it could also be subsonic or static.  When you see photos of shock diamonds from an engine test stand, it is obviously static.
The shock that emanates from the triple point and trails aft (from the red disk) is turning the flow back to parallel (notice the symmetry line nearby).  When that shock hits the ambient, if it were to continue, the pressure would not be equal.  Consequently, that shock reflects as an expansion fan, which turns the flow out.


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Does oscillation occur in the pitch axis of a plane?
If you a small GA plane, and apply full rudder to one side and then let go, there will be some oscillation from the momentum of the rotation about the yaw axis. Does this same effect happen with the pitch axis?
This can only probably happen when the plane is at a high angle of attack, but does it happen when maneuvering?
(Below is what I believe happens with the yaw axis in a sideslip, correct me if I'm wrong)

So say you have a fuselage without wings (still with tail section), fixed on its CoG. If there is a yaw disturbance, the V-stab will counteract that, sending the tail back to the original position. The thing is, once the tail is back where it was before the disturbance, there will still be the momentum of the tail section traveling in that direction, so it will continue past the original position and go the opposite direction of the way the disturbance made the tail section go.
Another way to think of it : Pretend you're in a sideslip because you're holding the rudder to where your nose is to the right of the direction of flight, in a smaller aircraft (just for simplicity). When you let go, the V-stab will counteract this sideslip, pushing the tail back to the right (viewed from behind the aircraft). When the tail arrives at the position where it is parallel to the freestream, it still has the momentum from getting pushed back from the sideslip.
That would make the tail now continue traveling to the other direction, creating some sort of oscillation as the process repeated itself, I think.

Now I don't see anything preventing this from happening in the pitch direction either. Am I missing something, or does it also happen in the pitch direction?

Yes it happens in pitch.  In fact, you generally learn about the pitch oscillations before the other modes.
There are two pitch oscillations that we generally talk about.  One is called the short period, the other is called the Phugoid.
The short period is most like what you're thinking of with yaw.  It is a fast oscillation that is quickly damped.  When a stable aircraft is perturbed from the trim condition (CM=0), there is a moment to correct.  That moment accelerates the aircraft back towards the equilibrium condition.  Once it gets moving, the rotational inertia causes the aircraft to 'over-shoot' the equilibrium condition, which causes the corrective moment back the other way, etc.  This motion is quickly damped out -- on the order of one second.
If you have a throwing dart, get someone else to throw the dart and stand to the side so you can see the flight path.  When they throw the dart straight,  you can't see anything special because the perturbation from equilibrium is so small.  Instead, have them throw the dart 90 degrees (or even 180) to the normal direction of flight.  You will be able to see the short period oscillation.
The other longitudinal oscillation is the Phugoid.  It is a very slow motion that we consider to happen at constant angle of attack (CL) and at equilibrium pitching moment.  It happens when the aircraft's trim airspeed is not the current airspeed.  It occurs as an exchange between potential and kinetic energy.  Each oscillation takes 30 seconds or longer and it can take minutes to damp out.  In normal circumstances, the pilot will cancel out this oscillation without thinking about it.
The source of damping in the Phugoid is aerodynamic drag, so as aircraft are designed to be very low drag, they will have less and less Phugoid damping.

Ah okay that makes sense, +1. So the Phugoid oscillation happens basically because the lift coefficient changes with airspeed, and when the aircraft is going a different speed than trimmed for, it will want to pitch. Once it has pitched, the airspeed will change, again changing the lift coefficient, and changing the flight path. Is this correct?

The lift coefficient is constant during the Phugoid.  When the airplane is going too fast, it will get too much lift, which will cause it to climb, losing airspeed - until it is too slow, then it will descend, gaining airspeed.  This is done at constant control inputs -- fixed trim conditions - but the aircraft is out of trim until the perturbation is damped and it gets back to equilibrium.


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Why do tip vortices seem to 'bend' inwards at the tip of a plane wing?
Why do tip vortices seem to 'bend' inwards at the tip of a plane wing?

Here if you look closely,  the tip vortices seem to bend or travel inwards very slightly after the aircraft has passed. In this picture, the effect is more exaggerated on the lower wing. What force makes the vortices bend inward?
My original thought was that the wake of the plane itself is the culprit, but I'm not sure, as the vortices seem to only bend right as they leave the wing, and then they travel straight.
It seems like if the wake was the cause then the vortices would continue to bend inward, and not just stop after the plane passed.

Because the back side of the wing is a large low pressure area?

@user3528438 Interesting thought, but right where the vortices bend isn't beside the wing, where the low pressure is. In other words, they bend after the area of low pressure on the wing, at least that's what it seems like. (I still could be wrong)

after the area of low pressure - remember that behind the wing is exactly where the wing was just a few moments ago so it is still an area of low pressure. Air does not teleport instantaneously back to ambient conditions when you're not there anymore. It takes time for air to move. This whole lag phenomena (inertia) is part of the whole reason vortices form in the fits place.

That is part of the wake rollup process.
The wake behind an airplane is caused by the downward acceleration of air while flowing around the wing. The tip vortices only show how the local streamline winds up in the wake, being made visible by local condensation. That the wake coming off the lower wing seems to "bend more inwards" is due to the downward motion of the air and the particular angle from which the photo was shot.
This answer goes into the details of wake formation. Note in particular that the distance between the cores of the trailing vortices is much more narrow than the wingspan. The air coming off the wingtips has been accelerated inwards on the upper and outwards on the lower side of the wing, and the resulting swirl causes low pressure at its core which in turn leads to the condensation that makes it visible.

What force makes the vortices bend inward?

It is important to not confuse the tip vortex with the full wake which forms behind the airplane. The downward moving air of the wake will suck in air from the sides and from above, and displace the air below it, so again a swirling motion will result. The condensed humidity of the tip vortex visualizes how this rollup evolves behind the wing. Local pressure gradients are responsible for the inward motion: This air tries to fill in the volume of lower pressure that is formed by the downward motion of the wake.

I added an answer with video clips of downwards moving tip vortices to help better understand what is going on.

What you see are two visual streamlines, which contract due to converting back to undisturbed airflow.
The forward flying aeroplane occupies space and pushes the air out of the way; when it is past, the air flows back to where the plane was. As simple as that. How fast, neat and turbulence free it flows back is a function of the shape of the physical body - how streamlined it is.

This answer makes no mention of the considerable effect that lift production has on the airflow around and behind an airplane. I would guess that in a zero-g maneuver at the same speed, the convergence of these streamlines would be considerably less (and I suppose one would have to use smoke to make the effect visible in that case.)

@sdenham -  I added an answer with video clips of downwards moving tip vortices to better understand what is going on (in a non-zero g case).

It might help to see video clips showing the downwards movement of wing tip vortices:
Plane flying through clouds

Owl gliding through bubble wall. The audio commentary is mis-leading, "downwards moving vortices ... helps provide lift". Clarifying: the vortices move downwards because of the downwash that coexists with lift. Although owls and some other raptor birds use their tails to generate lift more efficiently (less drag), birds like the Albatross rely on large wingspans (10 to 11 feet) for more efficient flight, and most gliders also rely on large wingspans for more efficient flight. The Nimbus 4 series rely on an 87 foot wingspan to achieve a 60:1 glide ratio at about 80 mph.

Brilliant videos - but the comments - aarghh! "Vortices spinning down from the wingtips" and "a tail that generated lift would make it unstable". Better to watch muted.

@PeterKämpf - I deleted my prior comments, and updated my answer to note the issues with the audio commentary in the second video.  My personal experience is limited to radio control gliders with relatively small tails and near zero lift from the the fuselage, and generally the CG is set relative to airfoils center of lift, since the effect from the tail is so small for these type of gliders which are setup with only a hint of positive pitch stability.  https://www.youtube.com/watch?v=ge_Ph96n_8Q

There's at least two things going on here.
As the engine's exhaust jet entrains surrounding air, the net  effect outside the entrainment region is that the jet contracts. The free stream wake contracts around it. That's what's happening well downstream of the plane.
There is also a bit of an inboard jump right behind the wing. That is due to real world engineering of sophisticated tip design and aileron settings. There are two vortex cores wrapped like a candy cane. The condensation happens to start at the tip with this aileron setting; and they roll up and combine a short distance behind the wing.


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Why are almost all winglets turned upwards?
Look at a 737, A320, A330, etc. All of them have upturned winglets. Why are they all upturned instead of downturned? Is there a reason for this or is it just how things ended up?
I was looking around and I found that the A-10 has downturned winglets, kind of (meaning not all winglets are upturned).

One of the reasons is probably due to the manufacturing complexity of the wings. Not sure though, as turning them upward probably causes the same amount of manufacturing complexity.

Newer 737NGs, 737 MAX, A300-600/A310, older A320s and the A380 have both up and down wingtip devices.

These are no winglets! Better use the absurdities which Boeing puts on its latest 737 models as an example - one part of them points a bit downwards.

One reason might be that upward winglets look better.  Don't they?

These wingtips are called Hoerner wing tips (they are not winglets).

Up winglets are less likely to hit the ground.
The clearance from the wingtip to the ground in a worst case situation is important.  Sometimes there is a requirement on the wingtip clearance when one main gear is at maximum extension and the other is at maximum squish with a flat tire.

Or a crosswind landing with an inconveniently timed wing dip that gets the aircraft rolling for some length of time before the pilot can counteract it.

Yes -- that is the sort of thing that would cause you to derive a requirement like this.

but they're not bigger than jet engines under the wing.  It must have some aerodynamic benefit

@user3413723 They're much further out than the engines...

Don't forget the servicing trucks.

Just to complement the other answers a bit.

I was looking around and I found that the A-10 has downturned winglets

Those wingtip devices on the A-10 are not really winglets but much simpler "dropped wingtips".
Even if they might improve the aerodynamic characteristics of the wing (especially maybe in ground effect), their main job is much easier i.e. protecting the ailerons and the chaff & flares from ground contact during low-level manoeuvring. The same trick is widely used in gliders. For the same reason also the vertical tailplanes protrude a bit downward in order to protect the elevators, especially during belly-landings (picture source):

And their use actually points us toward the answer to your question: virtually all winglets are turned upward (except the split-tip type) to avoid a dangerous contact with the ground.

It's important to separate winglets, which are thrust generators, from flow inhibiting fence-like devices like droop tips.
Winglets make thrust from the circulating flow, like a sail boat close hauled (Whitcombe originally called them "tip sails"). There is a lift force at 90 degrees to the spiraling flow mostly pushing the winglet toward the fuselage (or in the case of a sail boat, heeling it over), but angled a few degrees forward so there is a forward thrust vector. As a bonus, you have the downwash made by the winglet's airfoil which is redirecting the flow off its trailing edge outward, weakening the circulation.
You can theoretically extend it down, angle it slightly inboard instead of outboard so the lift force is out and forward instead of in and forward, and get a similar result, but you give up ground clearance if you want the winglet to be big enough to do any good.
The droop tip is a different device and is not making lift, just blocking some of the core circulation of the tip vortice. A winglet uses an airfoil to exploit the energy in the circulation flow to create to a counteracting force though lift generation, like any other wing, just horizontally.  Hence, wing-let.


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Why does the shock angle always seem to equal the "sound cone" angle?
(I'm going to use the term "sound cone" for the area you can hear an aircraft when it's at Mach ≥ 1, illustrated by this picture. Not considering the shockwave.)

So as explained in this answer, the angle that an oblique shockwave will form at is dependent on 2 things.

The Mach number before the shock
How much the flow has to turn in the shock

So this means that the shock angle will change with speed, but you can also change the shock angle by changing the aircraft shape (which would change the flow turning angle, therefore changing the shock angle).
The sound cone angle depends only on the speed, if I'm correct. When going Mach ≥ 1, the sound waves can't travel forwards relative to the plane because the plane is going faster than the sound, but they can travel sideways and back, evidenced by the picture above.
My confusion is here : Why does the sound cone angle always equal the shock angle?
(You can't hear an aircraft until after the shockwave)
Is it just a coincidence that the shock angle (determined by the flow turning angle and Mach number) equals the sound cone angle which only depends on the speed?
For an example (at Mach 1 in example) -- If you were to only change the nose shape of an aircraft, that would change the flow turning angle, and consequently the shock angle. You didn't change your speed, so the sound cone angle (again pictured above) wouldn't change. Wouldn't this mean at some point the sound cone and shockwave would meet, because they are not parallel? If they did intersect, the shock would travel through the sound cone and in that case you might be able to hear the aircraft before the shockwave.

My theory up to this point is that sound (say sound from a jet engine) can't travel through a shockwave, so it wouldn't matter if the shock and sound cone were parallel, because the sound can't travel through it anyway.

Because they are the same thing.  You are just using different words to describe the same thing.

@MichaelHall Hmm okay. So the sound cone angle is determined from the speed of the aircraft. The shock angle is determined by a different set of factors. Is that incorrect? (Sorry if this seems obvious)

(It might help to see where I’m coming from in the example from the second to last paragraph)

I will defer to the experts to explain how they are calculated if it's relevant, but what is the shockwave other than sound? Hasn't that been firmly established in the answer to one of your previous questions here?  Why wouldn't they be the same?

@MichaelHall Yeah I agree, but I always looked at it as if the shock was separate from the sound of the aircraft, if that makes sense. The shock is purely sound, but I don’t see why it’d be at the same angle as the sound cone. I could see how this might be a confusing question

@MichaelHall also do you know why this question is getting downvoted? I would improve it if I knew what was wrong.

I downvoted it because I thought you just weren't getting it and keep asking very similar questions on shock waves.  I thought the answer was obvious, but I am doubting myself now because Sophit is very smart on these matters.  However, I need a better explanation.  Anyway, I went to remove my downvote just now but it is locked in unless the question is edited...

Let us continue this discussion in chat.

Did not mean to do that, whoops

My confusion is here : Why does the sound cone angle always equal the shock angle?

Close to the body they don't. But further away they do.

The angle of the Mach cone is a pure geometric entity: using the picture in your question as basis, if you have a circle (the sound source) whose radius grows with a speed $a$ and which is moving with a speed $V$ bigger than $a$, then it "trails" along a inclined path which has an angle $\mu = \arcsin(\frac{a}{V})$ (by definition of $sin$). If $a$ is the speed of sound then $\frac{V}{a}=M$ (again by definition).

On the other hand, the angle $\beta$ at which an oblique shock wave forms upon encountering a wedge of angle $\theta$ is calculated applying the usual physical principles of continuity (mass cannot be created nor disintegrated) and momentum's conservation (second and third Newton's laws). What you get is the following equation which is normally solved via its plot (both pictures from Wikipedia):
$\tan \theta = 2\cot\beta\ \frac{M_1^2\sin^2\!\beta-1}{M_1^2(\gamma+\cos2\beta)+2}$

This plot is important because it helps us understand why the inclination of the shock wave changes.

Let's do a practical example and see what happens with our Concorde flying at Mach 2. Let's say that its nose has a (semi-) angle $\theta$ of 15°. So:

From the equation of the Mach cone we get an angle $\mu$ of 30°.
In the plot of the $\theta-\beta-M$ relation we follow the line labelled with a "2" (the Mach number) till it crosses with the vertical line at $\theta=15°$ (the nose of the Concorde). They intersect at $\beta=45°$ and $\beta=80°$ (both circled in red). The first solution is the so-called "weak" one while the second one is the "strong" solution. The "strong" shock wave normally happens in confined spaces (inlets for example) and we are not interested in it. Therefore our shock wave leaves the nose of the Concorde at 45° (point a).

The shock wave is therefore more "vertical" than the Mach cone (45° vs 30°) and should reach the terrain first.
But it doesn't.
Why that? Because "far enough" from the Concorde's nose, the path followed by the airflow is not affected by its presence and it's just a straight line. Close to the nose we need a shock wave at 45° to make the airflow turn the 15° of the nose. Anyway, a bit further, the flow turns a bit less than 15° (let's say 10°) and therefore a shock wave of 40° is enough to achieve this inclination (point b). Still a bit further, the flow turns again a bit less (let's say 5°) and therefore a shock wave of 35° is this time enough (point c). And still a bit further, we reach the "far enough" distance where the airflow is just going straight: we reached the point c where $\theta$ is zero (i.e. the airflow goes straight) and the shock wave is 30°. Those 30° are exactly the same 30° of the Mach cone!
The end.
To recap: approaching the pointy nose of the Concorde the airflow must turn 15° and this is achieved via an oblique shock wave at 45°. The further we are from the nose, the smaller the turn of the airflow is and therefore the smaller the inclination of the shock wave is. Far from the nose we reach the limit where the flow just travels straight on and the inclination of the shock wave coalesces into the Mach cone at 30°: the sonic boom starts as an oblique shock wave but ends as a Mach cone.

So an observer would hear a sonic boom, followed by a brief period of silence, and then hear the noise of the airplane?

@MichaelHall That’s what I was thinking of. Or as stated in my question, you could hear the noise of the airplane and then the sonic boom. This isn’t true, but I don’t know why, unless the very last paragraph of my question is correct.

@Wyatt, you most definitely would not hear the noise first and then the boom.  If you haven't realized this by now then you need to go back and do a thorough review of all answers to your previous questions on shock waves...

P.S.  Sophit, I completely trust your knowledge on this, but just throwing up equations without some context or narrative explanation doesn't answer the root question in a useful way for anyone other than a pure math geek.  Please elaborate for the rest of us.

@MichaelHall I realize that you would never hear the sound before the shock, but I don’t know why, as stated in my last comment.

@MichaelHall: I totally agree that my answer is thin, I'm expanding it with an example but I need some (free) time

@MichaelHall: done

Thanks for that edit @sophit, it helped a lot! So, sorry if I'm asking something already explained in you answer, but will the shock wave always be 'more vertical' than the Mach cone considering different plane geometry?

@Wyatt: yes always. The lines in the plot are always above the values where they meet the y-axis and those values correspond to the angle of the Mach cone. If you follow the line at Mach 2 of my example, you'll see that it meets the y-axis at 30°

@sophit, I think it's good to highlight that the first formula is a special case of the second formula. From far away, the influence of $\theta$ becomes insignificant.
If you look at the graph at $\theta=0$, for a given M it gives the same value for $\beta$ (for the weak shock), as the first formula gives for $\mu$

@ROIMaison: that's written at the end of my answer

@sophit, I'm not sure if you're saying the same as I am. I say that at the far-field, the value of $\theta$ is insignificant, and as consequence, the two become the same, whereas you say the shock wave is (and stays) more vertical and you'll hear the shock before you hear the engines.

@ROIMaison: Ah, now I get it. I've analysed only what happen close to the body while you mean what happens far from it. Yep right, I'll update the answer

@ROIMaison: done, thanks for the comment.

@Wyatt: Let me know if the explanation is now clearer.

@sophit Yeah it’s more easy to understand now, thanks. So I’ll say here what I interpreted it as: you have an object that makes the flow turn x degrees (makes it turn with a shock) and when that airflow leaves the influence area of the object, the shock angle doesn’t have to be as steep to get the same turning angle. Then (through some equations I can’t put here because a comment can only be but so long), the shock that has the less steep angle (aka the one outside of the influence area) will always equal the sound cone angle. Is this correct?

@Wyatt: yes correct, the Mach cone is the limit at which the shock wave tends

@sophit, great edit, it improved the answer a lot!

@ROIMaison: thanks! Your comment was important since I hadn't really got the real meaning of the question

Ah okay, thanks for the great answer! (Last question because I don't want to ask tons of questions) In the 15° turning angle case, the shock can be at a less steep angle further from the body. Example being the shock at 45° close to the body and 40° a little further away. If you were to change the flow turning angle to 10°, that would presumably change the shock angle values to maybe 40° close to the body and 35° a little further away. If you didn't change your speed and only the flow turning angle, the Mach cone would still stay the same (let's say 30° like in your answer). Wouldn't this mean

the shock would be at a lesser angle than the Mach cone, so you would hear the shock, then a brief silence, then the jet noise? Sorry if I'm missing something obvious because it feels like it lol. (Also I understand now why you can't hear a shock after the noise of the aircraft, the shock is noise and you can't hear noise before the mach cone)

The shock has always a steeper angle than the Mach cone: the point on the lines are always above the point where they intersect the y-axis (the Mach cone)

I know I literally just mentioned 'last question' in my last message, but I just had one more minor thing on my mind that I didn't feel warranted a separate post. In this answer, it says that the reason for the shock changing inclination as it moves away from the object is the low pressure behind the object. (See picture in the question).

I also know that the shock is always "more vertical" than the Mach cone, but in this scenario, say the low pressure somehow was able to bend the shockwave a lot to where it bent past the Mach cone angle. I can't see anything stopping the shock bending past the Mach cone angle. (Bending past meaning where you heard the aircraft and then the shockwave). In other words, it seems like the low pressure bending the shockwave would 'override' the other effects. Sorry for bugging you again!

Close to the body, the shock wave has an angle given by the oblique shock formula. Far from the body, the shock wave has an angle given by the mach cone formula. At the transition, the shock cone bends. See right photograph below for a very visible example of this bend.

Here's another diagram of the equations and terminology in sophit's answer.
(This isn't a proper answer as such.)

Haha love that diagram  It does give some good info.

The angle of the oblique shock wave is not gives so that a comparison cannot be done


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Where does the main CoP move when deflecting the elevator up & down and trimming?
I recently asked a question and in the comments I got this response
"The CP changes with angle of attack, so it is not a useful point in terms of stability. The NP does not change with angle of attack. Your intuition about distance from CM applies to the NP. When an aircraft is trimmed (at some alpha), the elevator has been adjusted such that the CP is at the CM."
My question was how does the CoP go to exactly where the CM is when trimmed? Wouldn't that mean the aircraft is far more unstable than if the CoP was still behind the CM by a bit?
(I didn't go to aeronautical engineering school or anything like that so if you could kind of put the response in layman's terms I'd appreciate that)

There's a lot of terminology there that must be well understood before proceeding: do you understand the meaning of Neutral Point, Centre of Pressure and pitching moment $C_m$ (I suppose of the wing)? For example: although only slightly, Neutral Point does change with AoA. And what's the "distance from CM"? There's no distance from a moment...

@sophit Here he is quoting me using CM to refer to the center of mass, not the moment coefficient.  I shouldn't have done it, but I think it matched the OP's usage in the original question.

@RobMcDonald haha I didn’t know that you’d see that. Thought this deserved a whole new question.

@RobMcDonald: ah ok, thanks for the clarification

The center of pressure is the point where the aerodynamic moments go to zero.  If you sum about this point, the pressures produce a pure force and zero moment.
If the center of pressure is not located at the CG, the aircraft will rotate.  Therefore, if an aircraft is trimmed, then the CP has been positioned to coincide with the CG.  We never think of it this way, but it has to be true.
*This argument ignores the moment caused by the engines.  And technically, the CP applies only to pressure, so it also ignores the moments caused by viscous shear forces.  But, the inviscid pressure forces dominate an aircraft -- they are what we use to hold it up (lift) and to control it (roll, pitch, yaw moments).
The CP changes with angle of attack -- and also with control surface deflection.  So, trimming an aircraft requires simultaneously finding the angle of attack and control surface deflections to achieve the desired lift coefficient (target trimmed airspeed) and zero moment.  Fortunately, for the most part, these equations are linear and we can solve this with a simple system of two linear equations.  Typically written as a 2x2 matrix equation.
Since the CP moves with angle of attack, it is not useful for the stability discussion.  Instead, that focuses on the aerodynamic center or the neutral point.
AC and NP are pretty much analogous terms.  We usually use AC when talking about 2D airfoils and also 3D wings in isolation.  We usually use NP when talking about a 3D aircraft as a whole.  The NP is a weighted (by area) average of the AC's of all lifting surfaces (wing, tail, canard) for an aircraft.
The AC and NP are defined as the point where the moment is constant as you change angle of attack.  This is only true in the linear, well-behaved region of flow.  When you approach stall (for example), this falls apart.

oh okay so basically the lift from the tailplane combined with the change in AoA of the wing will move the CoP to the CM?

@Wyatt Yes.  When a perturbation hits, a stable aircraft will return to the equilibrium point, and an unstable aircraft will diverge from the equilibrium point.

Ahh, okay makes sense. Thanks for all of your help!

Last question, say there is a pitch up disturbance, wouldn't that move the CoP forward, and because of a higher AoA making more lift forward of the CM, making the plane pitch up?

(In a stable plane with a restoring moment)

I never think in terms of center of pressure.  I do not find it to be useful.  I find that it adds more confusion than it is worth.  I encourage you to wash center of pressure from your mind and instead think in terms of aerodynamic center, neutral point, and center of gravity/mass.

Oh okay, so how might that work when you pitch up, considering the CoP is at the cm? Just is hard to not think about the cop because it’s where all the lift is made.

The lift is not made there.  The lift is made everywhere.  It is so much easier to think about a lift and a constant moment (that don't move) than it is to think about a lift and a zero moment that move around.

Oh alright. Makes sense, but doesn’t all of the lift total to the CoP?

What CoP.  I don't know what you're talking about.  That concept means nothing to me.  Drive it from your mind.

haha it’s very hard to. So if I don’t consider it where do i consider where the lift is made? How would the aircraft pitch up when a nose down disturbance happens or vice versa?

After letting that set into my mind, I think I know what you mean. Now when you define NP, are you talking about the whole aircraft’s np or just the wing (generally when talking about stability is what I’m asking)?

NP is for the whole aircraft.  AC usually applies to airfoils and wings in isolation.

Ah okay that’s what I thought. Thanks for all of your help, I appreciate it!


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Do winglets change the lift distribution of a wing?
In this answer, it claims that the reason for lift decreasing at the wingtips is due to airflow around the wingtips. If I'm thinking correctly, that is due to the air below the wing spilling on to the top, decreasing lift (correct me if I'm wrong). That made me wonder if adding winglets evens out the pressure distribution along the wing, because less air spilling over (from bottom of wingtip to top) would increase lift, right? In that case, say you went to an extreme, and made a massive winglet, would that basically get rid of tip vortices and make the pressure distribution almost completely even?

A closed wing would be even better.

@sophit Closed wings have vortices still, correct? Would there still be air leaking from the bottom of the wing to the top in that case?

Does this answer your question? Is a winglet better than an equal span extension?

Winglets do change the lift distribution on a wing.
To really take your line of thought to an extreme, replace the winglets with flat walls that extend up, down, forward and aft to infinity.
If you place a rectangular wing in a wind tunnel such that the span is equal to the width of the wind tunnel, you can simulate two-dimensional flow with a uniform lift distribution across the wing.
Most high quality airfoil data has been taken in a tunnel set up to use this effect.
Of course, winglets also add weight and wetted area (skin friction drag).  So you don't want to install a giant winglet on an aircraft.

ah, yeah that's what I was thinking. I was a bit confused because I just assumed there was another factor that also causes it. Thanks for your help!

Whitcombe originally called winglets "tip sails", because they operate much like sails on a boat. They exploit the energy in the tip circulation by placing an airfoil surface such that lift is generated inboard. Since the circulation flow is moving inboard, it has a positive angle of attack to the winglet and the lift force is perpendicular to that angle of attack. The lift force is pushing the winglet toward the fuselage, but also angled forward somewhat. The forward component of the lift force is, literally, thrust as far as the airplane is concerned. Winglets make thrust, like a sail boat close hauled.
While making that thrust they are consuming the energy in the circulation and creating an outwash (downwash sideways) that opposes the circulation, weakening the core of the vortice and having an effect similar to a span increase.
The end result is the lift distribution is improved similar to a span increase, and some thrust benefit is also extracted from the flow that may exceed the winglet's drag.
They only generate enough energy to compensate for their own drag when the wing is "working hard" operating at higher AOAs, that is, low indicated airspeeds. The only airplanes that cruise at low indicated airspeeds are jets that fly above 30000 ft (for example, 443kt TAS at 37000ft is only 250 kt indicated). This is generally why you only see winglets on jets, and putting them on straight wing low altitude aircraft is mostly for looks, or to improve rate of climb (where AOA is relatively high enough to get useful work out of them).

besides being only beneficial at high AoA, a winglet also helps to stay in a lower wingspan class. Only then is it really justified.

I'd always assumed the funny turned up tips on the Canadair CL-415 were some kind of winglet intended to improve its climb performance. An aerodynamicist chum who had some involvement with the program told me they were actually a bandaid to restore dihedral effect that was lost due to the tall, flat fin-like PW120 nacelles, that replaced the round R2800 nacelles, making a high wing configuration into effectively a quasi mid wing configuration, in a plane with no geometric dihedral.

Same thing with the turned-down wingtips of the He-162. They were a band-aid for too much dihedral and since the design had to be ready for service soon, there was no time to change the jigs to a smaller dihedral. They were called "Lippisch-Ohren" (Lippisch ears) after the person who used them first on his tailless designs.

I've read that after the kinks were worked out with that, and increasing vertical stab size IIRC, they were actually very nice handling aircraft.

Yes, a later version also had a longer tail which improved handling a lot.

@PeterKämpf Off topic slightly: My homebuilt, a Pazmany PL-2 has 5 deg of dihedral. increased from 3 deg in the PL-1.  But the yaw-roll couple is surprisingly weak. I need to put in full rudder and hold it to get any kind of roll rate.  I would never be able to maneuver the plane in a useful way using rudder alone.  Not sure if it's the wing tip pod fuel tanks are affecting things, the dihedral angle is too small, the relatively long span (27 ft) is a factor, or all of the above.

I don't think the tip tanks are at fault. After all, the maximum sideslip angle should be less than 15 degrees. But since low wing airplanes need more dihedral, those 5° are still too low.

“The lift force is pushing the winglet toward the fuselage, but also angled forward somewhat.” Wouldn’t this also require the winglets to be at an angle? If you were looking at the plane from a top down view (with the plane facing forward) the winglets would need to be facing slightly outward, right? If so, wouldn’t having the winglets at an angle induce drag of its own? (Sorry for the late comment; I just came back to this answer to re-read it)

Depends on the local angle of attack of the circulating flow. You often see a bit of toe out of the winglets to optimize the AOA. On the CRJs this was obvious just standing under them, about 2 degrees from the looks of it. OTOH I fly an ASW 24 glider from time to time and its winglets don't appear to have any toe-out; they look straight, but have a cambered airfoil contour. The ASW's winglets plug in with pins and are just held in place with tape, the inboard lift forces on them doing most of the holding in place.


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Will air in a shockwave flow outwards?
It's well known that shockwaves extend outward from a plane, as seen in this picture.
My question is, does the high-pressure air in the shockwave extend outward? This is hard to explain so I'll let the picture do the work.

Basically I'm asking if air will flow where the black arrow is pointing, as the shock is high pressure and I don't see anything stopping it. I know that the shockwave won't go past the supersonic pocket of air, but will the air in the shockwave flow in the vector the arrow is pointing at?
Imagine that the shock was just a high pressure, thin line of gas and nothing more. The air would try to flow in the direction of the arrow, right?

In reality, the air isn't going Mach 0.8, the airfoil is.  The accelerated flow runs into the boundary layer attached to the airfoil.  This is where the shockwave forms.  Air may indeed be deflected in the direction of the arrow, but the normal shockwave may not be stable at all.  Early pioneers of this flight regime described shockwaves as "dancing across" the wing surface.  Area ruling was developed to literally give the compressed air somewhere to go.

No.
The shock is not high pressure.
The shock is a jump in properties.  The flow is higher pressure behind the shock than in front.  Think of the shock as an artificial boundary where the properties change.
Like the place where water meets air -- that interface is a rapid change in properties.
In the case of your picture, the shock does not turn the flow.

Oh okay interesting. Would you mind telling me some of the properties that change?

Mainly pressure, density, and temperature.

One of the fundamental principles of aerodynamics is that particles of air remain in their streamline. This is not strictly true in turbulent error where air is mixed. But it is an important aspect of understanding the answer.

@CamilleGoudeseune Does the pressure go up or down? (compared to ambient)

I think I understand it more now. So basically the transition of properties happens over the shockwave and not instantly, so the high pressure doesn’t happen much until after the shockwave? Shockwaves are also very thin, so I’d imagine the air wouldn’t have much time to flow outwards like my question was talking about.

Rob McDonald, you should get payed by @Wyatt

The flow velocity vector can also change at a shock (according to the Rankine-Hugoniot conditions). The velocity does not change at a contact discontinuity. On a shock it generally does change and so it does in a rarefaction wave.

@Wyatt Which high pressure are you talking about? High pressure happens on the pressure side of the airfoil. On the suction side we have a low pressure. After the  shock wave or after the flow separation the pressure is some what higher, but still not really very high, otherwise you would lose the lift. Check, for example, https://www.grc.nasa.gov/www/wind/valid/raetaf/raetaf04/raetaf04.html especially figure 7 and figure 8. fig. 8 maybe harder to understand at first so fig. 7 is easier.

@sophit haha I would! He has been very helpful!

@VladimirFГероямслава Hmm okay, interesting. I for some reason always thought it was much higher pressure behind the shock. Thank you for those airfoil plots!

It's not the air in the shockwave flowing in the direction of the vector, it's the energy being propagated in the form of a wave.
Waves transfer energy through a media.  The water in a wave that travels across an ocean does not come from it's source, only it's energy.
Waveforms can hold their shape a long distance.  One can transfer energy from you voice into paper cup onto a string, and the string will vibrate for many feet and can be heard as sound by attaching it to the bottom of another paper cup.
Shorter wavelengths, such as visible light, can carry even greater distances.
Sonic waves created by an aircraft reaching Mach 1 can be 120 decibels or louder.  These sound waves, in a similar range to thunder, can carry for miles.

The physics of light don't apply directly to aerodynamics. I don't think mentioning them is helpful here. Additionally, the shockwave is a nonlinear boundary in wave behavior. So almost everything you say about waves does not hold equally on both sides of the shock wave. In fact, sound stops propagating normally at the shockwave.

@MarkJonesJr.  Well, radio waves apply to aviation, don't they?  What I am trying to illustrate, as the OP has asked multiple questions about this subject, is that wave energy is not about movement of one parcel of air in the direction the sound is traveling.  The energy (that creates "sound" against our eardrums) is transferred through air as a high and low pressure "pulse".  Correct me if I'm wrong:  through the same medium speed of sound does not diminish over distance, amplitude of the pressure pulse does.

@MarkJonesJr. "Sound stops propagating normally at the shockwve".  Would it hold that the normal shockwave ends where the air flow is no longer supersonic?  The OP has the shockwave extending past it.  Perhaps you can help with an answer.  Does the freestream disrupt this wave, preventing it from propagating?  I'm seeing airflow (and turbulence) over the wing aft of the shockwave.

authors: Vladimir F Героям слава


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How does aspect ratio affect stall speed and stall AoA?
I've read a few answers on here but they didn't really tell me what I wanted to know. So I get that high AR wings will stall before low AR wings but what makes that true? Higher AR wings have a higher lift curve slope (and from what I understand that means for a given change in AoA, the wing will create more or less lift), so wouldn't that mean for a low AR wing that has a lower lift curve slope that it will make less lift at high AoA compared to a high AR wing at a high AoA, and making a stall happen?
I basically am asking why there is a direct relation between AoA, AR, and stall speeds. Also, because I don't think I have a good understanding of the lift curve slope, if someone could explain that, I'd really appreciate it.

I don't think I have a good understanding of the lift curve slope

As much as I normally abstain from using potential flow theory to explain things, for this purpose it is actually useful.
In potential flow theory, lift can be calculated as the linear superposition of a contribution from camber and one from angle of attack. While the camber-related part of lift is constant, the angle-of-attack related part varies linearly with this parameter. Think of a wing as a vane which re-directs the flow of air slightly downwards: The higher the angle of attack, the more downward velocity is imparted on the air. Looking at the impulse change caused by this redirection we find a force, the resulting aerodynamic force. This can be split into a component orthogonal to the original direction of flow, called lift, and a component parallel to it, called drag.
As long as flow stays attached to the surface of the wing, the streamlines and flow patterns look very similar to potential flow, except for a small layer around the surface, called the boundary layer. Once separation sets in, the linear relation between lift and AoA breaks down. The variation of lift over the full range of AoA is shown in this answer.

I get that high AR wings will stall before low AR wings but what makes that true?

Since the pressure difference between the lower and upper side of a wing can equalize at the tip, local lift drops the closer you move to the tip. This tip effect is larger in low aspect ratio wings simply because more of the wing area is close to one tip. This means not only that a low aspect ratio wing needs more AoA to create the same lift as a high aspect ratio wing, but also that the pressure gradients over it are more shallow at the same AoA. Therefore, separation needs more AoA to develop on a low aspect ratio wing and it will reach a higher AoA before stall.
Still, the tip effect will reduce the total amount of lift the wing is capable to develop. If you look at the local lift distribution of a rectangular wing over span, an increase in aspect ratio will make it look fuller. The decrease in lift at the tips will affect a relatively smaller part of the wing while the part away from the tip will show an almost constant lift over span. In other words: The wing is working harder, so the stall speed of an airplane with a high  aspect ratio wing will be lower than that of a comparable airplane with a low aspect ratio wing. Only the vortex developing on highly swept leading edges will reverse that trend.

From the comments:

how exactly do leading edge vortices affect stall speed of higher and lower AR wings?

The vortex only forms at high sweep, which prohibits a high aspect ratio. By adding a rotational speed component to the air on the upper side of the wing, the pressure there drops further, adding more lift and lowering stall speed. Of course you want this low pressure to work on as much area as possible, so the inner chord of a wing with voretx lift should be high. Which invariably leads one to a delta wing. And deltas only come with a low aspect ratio.
Typical stall angle of attack of a high aspect ratio wing: 10° - 12°
Typical stall angle of attack of a low aspect ratio wing: 15° - 18°
Typical stall angle of attack of a delta wing with vortex lift: 25° - 30°

Ah okay thanks for your answer. One question I have is that how exactly do leading edge vortices affect stall speed of higher and lower AR wings?

Leading edge vortices are only a thing for highly swept, very low AR wings.  Typically delta wings, but they don't have to be a delta configuration.

@RobMcDonald ah okay but aspect ratio doesn’t affect the pressure difference between the lower and upper surfaces of the wing, correct? (Maybe a little because they can make more lift)

@Wyatt I said nothing about that.  Yes AR affects the pressure distribution -- and when you integrate it, the load distribution.  High AR wings are better than low AR wings -- they have higher lift curve slope and higher CLmax than low AR wings.
That has nothing to do with leading edge vortices.  LE vortices are only a thing for highly swept, very low AR wings.

(I meant high aspect ratio wings can make more lift)

@RobMcDonald ah okay. One of my original questions was why the low AR wings are the only ones with leading edge vortices. Also, I don’t want to bore you with questions, but how might AR affect pressure distribution and load distribution?

@Peter Ah okay that edit clarifies things for me a lot, thanks!

@Wyatt  It isn't so much that low AR wings are the only ones with LE vortices.  It is that only highly swept (and sharp) LE (45-60 deg) have LE vortices.  This amount of sweep is totally impractical to make into a high AR wing for many reasons.  Consequently, we typically only see that much LE sweep in fighter and delta wings (or possibly leading edge extensions) with AR less than 2.0.

@Wyatt for a straight, un-tapered wing, increasing the Aspect ratio mostly stretches out the load distribution and increases the region of nearly 2D flow in the center of the wing.  You should really just get your hands on a VLM or LLT tool (or write your own) and run some cases to play around.  Try VSPAERO, AVL, XFLR5, Tornado, etc.

@RobMcDonald ah, I see. I’ll check some of those pieces of software out for sure, thanks!

I don't believe this trend (higher CLmax for low AR wings) is true for straight, un-tapered wings.
Instead, low AR wings are usually used on high performance (fighter) aircraft, which have highly swept leading edges -- and relatively thin airfoils with relatively sharp leading edges.  These kinds of wings are able to set up strong leading edge vortices that can substantially delay stall and increase CLmax.
The Concorde is a great example of putting this phenomena to great effect.  The drooped nose of the Concorde is there to provide runway visibility during landing when the aircraft is at very high AoA.


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Does separation happen easier with less dense air?
Does less dense air affect when separation happens?

So if you have a cylinder traveling in a straight path like the one above, there is separation happening at a given speed. If you were to change only the air density, would this separation be less at the same speed, the same, or more? If it is different, why?

The Reynolds number is
$$Re=\frac{\rho\, u\, L}{\mu}$$
If the density $\rho$ were to be less, you would have a lower Reynolds number (all else equal).
In general, that would tend to make the separation happen earlier.
However, it really depends on where you are starting -- is the flow laminar or turbulent.  Perhaps you're at extremely low Reynolds number -- creeping flow even.
Whatever happens, it is determined by the Reynolds number.

I see. So to summarize, the lower density air will have a lower Reynolds’s number (more laminar).  Since laminar flow will separate before a turbulent flow, that would mean less dense air will separate easier?

Yes, I said in general because this is what happens for the range of Re we usually think about for airplanes and such.  At extremely low Reynolds number (less than 200), flow takes on a different character.  At even lower Reynolds number (much less than 1), flow takes on even more different character.  It is still laminar in these regions, but it is extremely viscous (think glycerine or molasses).  At these Re, the flow might not separate at all.

The ubiquitous plot of the drag coefficient of a cylinder in respect to the Reynolds number is worth a thousand words (plot source):

As you can see, at higher Reynolds numbers (i.e. at higher density) the drag coefficient becomes smaller.

Does separation happen easier with less dense air?

This is represented by the jump in the coefficient at around Re $3 \times 10^5$: here we have the transition of the boundary layer from laminar to turbulent. A turbulent boundary layer remains attached to the surface a bit longer than a laminar one and this results in a lower drag.
This is even better represented in the following plot (source - many thanks to @ROIMaison) which shows at what angle on the cylinder's surface the flow detaches:

Here 0° means the side of the cylinder invested by the flow, 90° its crest and 180° the opposite side. Also here it can be seen that beyond the Re values as given before, the (turbulent) boundary layer remains attached till higher angles (around 120°) than at lower Re (around 80°).
So yes, separation (and the relevant higher drag) actually happens easier with lower dense air (left part of the plot).

Well, this is true at least for all Reynolds numbers of practical interest. Anyway, when Re becomes very small (smaller than one - so called creeping flow), the flow crosses the body following its shape without separation (180°), as can be seen for example in the following picture taken from the legendary "An Album of Fluid Motion" by "Milton Van Dyke" (pdf here):

It is a good answer, but I think the reasoning becomes clearer when you show separation angle vs Reynolds numbers (such as this one).
That way you can answer the question directly, without making the 'detour' to drag.

@ROIMaison: uh, that's a nice plot, can I update my answer with it? Thanks for the editing

Yes of course, for reference; here is the link leading to the publication that contains the image

@ROIMaison I had a separation angle vs. Re plot in my mind when I wrote my response, but I couldn't find one.  I didn't include a C_D plot because (while related) it is a different beast.  It is too bad that chart doesn't go down to really low Reynolds number (1000, 100, 10, 1 etc) -- in creeping flow, you have 100% attachment.

@RobMcDonald, I also found a plot for the low reynolds numbers. It's the first image of the publication, ResearchGate it looks a bit odd. You can find it here

@RobMcDonald, (comment got locked). I agree that CD is an aggregate of various effects, and is a bit noisy and that this is better avoided by using the RE - separation plot directly. The plot above nicely shows the low Re asymptotic behavior.

@ROIMaison Great find.  That supplies data at Re from 7 to 200 -- and the other chart goes down to 4000 -- hopefully nothing too interesting happens in-between there (I think we're OK).

@ROIMaison: thanks, I've update my answer with your plot

@RobMcDonald: thanks for the link. I let you put your interesting plots about what happens at very low Reynolds numbers in your own answer

@sophit  Although all the definitions of fluid dynamic ranges vary based on who you talk to, I think that creeping flow is typically Re << 1.0 -- very small indeed.  I agree it doesn't apply to Aviation in the external flow sense, but these super low Reynolds numbers are important for things like thin film lubrication.

@RobMcDonald: aren't that Stokes flow? I updated the answer, thanks again

I'm not sure of an authoritative definition of creeping flow -- the Wikipedia article for creeping flow redirects to Stokes flow.  Your image from album of fluid motion has a Re=0.16.  I just had the feeling that by Re=1000, you'd find yourself in another regime.


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Why are bow shocks angled slightly downward at the leading edges of wings?
Why are bow shocks angled slightly downward on airfoils? In this picture if you look closely at the very bottom airfoil diagram, you'll see the "subsonic airflow" part is more below the leading edge, or the blue part extends more down than up. Very slight, but it's visible. Other diagrams have this too, why is that?
(also little side question, why are there lines that look like a bow shock on the bottom airfoil at the trailing edge?)

I wouldn't take this diagram as being scientific or to-scale.  It is an artists rendition of a diagram that has been copied and repeated many times, some small details get lost in translation.
In this diagram, in the first two charts, the blue region represents the area of supersonic flow.  However, in the third diagram, they've switched meaning and the blue area is subsonic flow.  Very tricky.
The only place a shock wave will be perpendicular to the flow is along the stagnation streamline.  All the other streamlines, the shock helps to turn the flow, and it takes an angled shock to turn the flow.
There is a second set of shocks at the trailing edge of the airfoil.  The bottom surface trailing edge flow is about horizontal, and the upper surface TE flow is down about 10deg.  After the trailing edge, these two must be parallel (and the pressures equal).  So, the top flow will turn a few degrees up, and the bottom flow will turn a few degrees down.  Since the flow before the trailing edge is supersonic, it requires a shock wave to make this turn happen.

Ah okay that clears things up, thanks. So in the first 2 diagrams, do those shocks disappear at supersonic speeds, or do they happen along with the bow shock? Also, would the correct terminology for the trailing edge shock be an oblique shock?

They are the same shocks.  The first weak shock appears at some subsonic Mach number.  It moves aft to the trailing edge as the region of supersonic flow grows.  Then, as you reach Mach 1, the bow shock forms.  Yes, I would consider the trailing edge shocks to be oblique shocks.

Oh okay, thanks a lot. So that would mean that the shocks in diagram 1 and 2 don't disappear, right?

Yes, they are the same shocks.  They move aft until they reach the trailing edge.

I see. Last question, do those trailing edge shocks happen at sub-mach speeds?

They happen when the region of supersonic flow above the wing reaches all the way back to the trailing edge.  It is probably possible for that to happen at M<1, but in normal circumstances that would be something like M=0.98 or so.


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Why does a slip line form in between these 2 shocks?

As seen here, the trailing edge shock creates what looks like a slip line between the two shocks. I thought a slip line only formed when air that went through different shock systems interacted, so what am I missing?
Why does this slip line form in this case?

I think this is just be the boundary layer leaving the trailing edge

Either this wing is at a slight angle of attack -- resulting in different accelerations across the uppper and lower surface -- resulting in different Mach number before the shock and therefore different shocks -- and therefore a slip line.
Or, the shocks are identical and what you are seeing is not a slip line, but instead the trace of the boundary layer.
I am not positive the technique used to generate this image.  It may be a Schlieren photograph, shadowgraph, or inteferogram.  Each of these will render different flow features slightly differently, but they're fundamentally based on visualizing gradients in density.

Ah okay that makes sense. In this case, do you think the boundary layer is higher density, so it shows up on this picture? I thought the boundary layer was the same pressure/density as the surrounding air, but just more turbulent in some cases.

Pressure is constant across a boundary layer, but temperature can vary.  P=rho R T.  These techniques image density gradient -- so changes in density, either increasing or decreasing.

Ah okay, thanks! I imagine the temperature is slightly higher in the BL, as it's experiencing the most friction. Also, can you model the pressure in the BL as $P=pRT$?

P=rho R T works as long as a gas behaves like a gas.  When a gas starts acting like a plasma (extreme high temperatures and low pressures) or like a liquid (extreme high pressures and low temperatures) this can break down.  You have to be careful in chemically reacting flow because the value of R can change -- but P=rho R T still works.

I see. So is $P=pRT$ the same as P=rho R T? (I'm not too familiar with P = rho R T)

rho is the spelling of the greek letter $\rho$, which is the symbol for density.  The symbol you are using looks to me like a lower case  p rather than a $\rho$.

Ah okay, thanks for the clarification!